Same solution... some more details

a1 + a2 + a3 = 21 ----> (1)
a2 + a3 = 18 -----> (2)

From Eqn (1) & (2),
a1 = 21-18 = 3

Now, since each term after the first equals r times the previous term, we have:
a2 = a1 * r
a3 = a2 * r = a1 * r^2

Plugging the values into a2 & a3 in Eqn (2)
a1*r + a1*r^2 = 18
3 * (r + r^2) = 18
r(r+1) = 6

r^2 + r - 6 = 0
(r - 2)(r + 3) = 0
r = -3 or 2
So we know it must be 2.

Then we can solve the question:

a3 + a4 + a5 = a1 * (r^2 + r^3 + r^4)
= 3 * (4 + 8 + 16)
= 3 * 28
= 84